A bath can be filled by the cold water pipe in 10 minutes and by the hot water pipe in 15 minutes. A person leaves the bathroom after turning on both the pipes. He returns just when the bath should have been full. Finding however, the waste pipe was open, he closes it. In 4 minutes more, the bath is full. In what time will the waste water pipe empty it?

### Concept & Equivalent Work Volumes If an expected outcome isn't met due to a leak, the "extra" work required to fix the deficit equals exactly the amount of work the leak undid during the time it was active. $$ \text{Volume Emptied by Leak} = \text{Volume Refilled to Catch Up} $$ ### Step-by-Step Solution 1. **Given**: Cold pipe (A) fills in 10 min. Hot pipe (B) fills in 15 min. Waste pipe (C) is open initially. 2. **Calculation**: - Total capacity = LCM(10, 15) = 30 units. - Efficiency A = 30 / 10 = 3 units/min. - Efficiency B = 30 / 15 = 2 units/min. - Normal combined efficiency (A + B) = 3 + 2 = 5 units/min. 3. Determine expected time of return: - Expected time to fill 30 units = 30 / 5 = 6 minutes. - The person returns after 6 minutes. During this time, the waste pipe was open. 4. Analyze the catch-up phase: - He closes the waste pipe, and A + B fill the tank in 4 more minutes. - Volume filled in these 4 minutes = $4 \text{ min} \times 5 \text{ units/min} = 20$ units. - This means exactly 20 units were missing at the 6-minute mark. 5. Calculate waste pipe efficiency: - The waste pipe emptied these 20 units during the initial 6 minutes it was open. - Efficiency of waste pipe = 20 units / 6 min = 10/3 units/min. 6. Determine time to empty full tank: - Time = Total Capacity / Waste Efficiency = $30 / (10/3) = (30 \times 3) / 10 = 9$ minutes. ### Exam Strategy & Shortcut Equate the works directly: The work done by the waste pipe in 6 minutes is equal to the work done by the filling pipes in 4 minutes. So, $6 \times E_{\text{waste}} = 4 \times E_{\text{fill}}$. Since $E_{\text{fill}}$ takes 6 minutes for the whole tank, the whole tank is $6 \times E_{\text{fill}}$. We need time $T = \text{Total} / E_{\text{waste}}$. Substitute $E_{\text{waste}} = \frac{4}{6} E_{\text{fill}}$. Time = $6 / (4/6) = 36/4 = 9$ minutes. ### Common Pitfall A common pitfall is calculating the net efficiency with all three pipes and creating a complex equation, when simply equating the leaked volume to the make-up volume is much faster and clearer. ### Final Answer Therefore, the correct answer is **9 minutes**.